Ring theory MOC

Monoid ring

Let 𝑅 be a ring and 𝑀 be a monoid¹. The monoid ring 𝑅[𝑀] is the extension ring of 𝑅 by adjoining 𝑀 in the most general way maintaining the monoid product as ring multiplication, ring as formalized by the Universal property. Thus it is an R-monoid constructed from the free module 𝑅(𝑀).

Universal property

Let 𝑅 be a ring and 𝑀 be a monoid. The associated monoid ring is a triple consisting of a ring 𝑅[𝑀], a ring homomorphism 𝜄 :𝑅 →𝑅[𝑀], and a monoid homomorphism 𝜇 :𝑀 →𝑅[𝑀]; such that given any ring 𝑇, ring homomorphism 𝑖 :𝑅 →𝑇, and monoid homomorphism 𝑚 :𝑀 →𝑇 there exists a unique ring homomorphism 𝑓 :𝑅[𝑀] →𝑇 such that the following commutes in \Set

This admits a unique extension to a bifunctor ( −)[ −] :𝖱𝗂𝗇𝗀 ×𝖬𝗈𝗇 →𝖱𝗂𝗇𝗀 such that

𝜄:Π1⇒(−)[−]:𝖱𝗂𝗇𝗀×𝖬𝗈𝗇→𝖱𝗂𝗇𝗀𝜇:Π2⇒(−)[−]:𝖱𝗂𝗇𝗀×𝖬𝗈𝗇→𝖬𝗈𝗇

become natural transformations.

Construction as maps

As with the free module, 𝑅[𝑀] may be constructed as the set of maps of finite support 𝑀 →𝑅, where we identify 𝑚 ∈𝑀 with 𝜇(𝑚) =𝛿𝑚 :𝑡 ↦[𝑚 =𝑡] invoking an Iverson bracket, and elements of 𝑅 with constant functions. For 𝑎,𝑏 ∈𝑅[𝑀], the product is given by

[𝑎∗𝑏](𝑥)=∑𝑚𝑛=𝑥𝑎(𝑚)𝑏(𝑛)

Proof of universal property

Clearly 𝑅[𝑀] is an abelian group under pointwise addition. The convolution operation is associative, since

[𝑎∗(𝑏∗𝑐)](𝑥)=∑𝑚𝑛=𝑥𝑎(𝑚)[𝑏∗𝑐](𝑛)=∑𝑚𝑛=𝑥∑𝑘ℓ=𝑛𝑎(𝑚)𝑏(𝑘)𝑐(ℓ)=∑𝛼𝛽𝛾=𝑥𝑎(𝛼)𝑏(𝛽)𝑐(𝛾)=∑𝑚𝑛=𝑥∑𝑘ℓ=𝑚𝑎(𝑘)𝑏(ℓ)𝑐(𝑛)=∑𝑚𝑛=𝑥[𝑎∗𝑏](𝑚)𝑐(𝑛)=[(𝑎∗𝑏)∗𝑐](𝑥)

and a multiplicative identity is given by 𝜄(1) =1. Clearly 𝜄 is a Ring monomorphism, and 𝜇 is a monoid monomorphism since

[𝛿𝑚∗𝛿𝑛](𝑥)=∑𝑘ℓ=𝑥𝛿𝑚(𝑘)𝛿𝑛(ℓ)=𝛿𝑚𝑛(𝑥)

Now suppose 𝑇,𝑖,𝑚 is another such triple. For the diagram to commute, we require that 𝑓(𝑟) =𝑖(𝑟) for all 𝑟 ∈𝑅 and that 𝑓(𝛿𝑛) =𝑚(𝑛) for all 𝑚 ∈𝑀. For 𝑓 to be a ring homomorphism, it follows

𝑓(𝑟𝛿𝑛)=𝑖(𝑟)𝑚(𝑛)

and thus for 𝑎 ∈𝑅[𝑀]

𝑓(𝑎)=∑𝑛∈𝑀𝑖(𝑎(𝑛))𝑚(𝑛)

which is unique, as required.

See also

• Group ring

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Footnotes

1. Or a semigroup, where one simply uses its completion to a monoid. ↩